Internal
problem
ID
[8171] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
34 Date
solved
:
Sunday, November 10, 2024 at 08:19:58 PM CAS
classification
:
[_quadrature]
\begin{align*} p \left (\operatorname {t0} \right )&=\operatorname {p0} \end{align*}
Existence and uniqueness analysis
This is non linear first order ODE. In canonical form it is written as
\begin{align*} p^{\prime } &= f(t,p)\\ &= -b \,p^{2}+a p \end{align*}
The \(p\) domain of \(f(t,p)\) when \(t=\operatorname {t0}\) is
\[
\{-\infty <p <\infty \}
\]
But the point \(p_0 = \operatorname {p0}\) is not inside this domain. Hence existence and
uniqueness theorem does not apply. There could be infinite number of solutions, or one
solution or no solution at all.
Solved as first order autonomous ode
Time used: 0.665 (sec)
Integrating gives
\begin{align*} \int \frac {1}{-b \,p^{2}+a p}d p &= dt\\ \frac {-\ln \left (b p -a \right )+\ln \left (p \right )}{a}&= t +c_1 \end{align*}
The following diagram is the phase line diagram. It classifies each of the above
equilibrium points as stable or not stable or semi-stable.
Solving for the constant of integration from initial conditions, the solution becomes
\begin{align*} \frac {-\ln \left (b p-a \right )+\ln \left (p\right )}{a} = t +\frac {-\operatorname {t0} a -\ln \left (\operatorname {p0} b -a \right )+\ln \left (\operatorname {p0} \right )}{a} \end{align*}
Solving for \(p\) gives
\begin{align*}
p &= \frac {a \operatorname {p0}}{-{\mathrm e}^{-t a +\operatorname {t0} a} b \operatorname {p0} +{\mathrm e}^{-t a +\operatorname {t0} a} a +\operatorname {p0} b} \\
\end{align*}
Summary of solutions found
\begin{align*}
p &= \frac {a \operatorname {p0}}{-{\mathrm e}^{-t a +\operatorname {t0} a} b \operatorname {p0} +{\mathrm e}^{-t a +\operatorname {t0} a} a +\operatorname {p0} b} \\
\end{align*}
Solved as first order Bernoulli ode
Time used: 0.280 (sec)
In canonical form, the ODE is
\begin{align*} p' &= F(t,p)\\ &= -b \,p^{2}+a p \end{align*}
This is a Bernoulli ODE.
\[ p' = \left (a\right ) p + \left (-b\right )p^{2} \tag {1} \]
The standard Bernoulli ODE has the form
\[ p' = f_0(t)p+f_1(t)p^n \tag {2} \]
Comparing this to (1)
shows that
\begin{align*} f_0 &=a\\ f_1 &=-b \end{align*}
The first step is to divide the above equation by \(p^n \) which gives
The next step is use the
substitution \(v = p^{1-n}\) in equation (3) which generates a new ODE in \(v \left (t \right )\) which will be linear and can be
easily solved using an integrating factor. Backsubstitution then gives the solution \(p(t)\) which is
what we want.
This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows
that
\begin{align*} f_0(t)&=a\\ f_1(t)&=-b\\ n &=2 \end{align*}
Dividing both sides of ODE (1) by \(p^n=p^{2}\) gives
Substituting equations (5) and (6) into equation (4) gives
\begin{align*} -v^{\prime }\left (t \right )&= a v \left (t \right )-b\\ v' &= -a v +b \tag {7} \end{align*}
The above now is a linear ODE in \(v \left (t \right )\) which is now solved.
Integrating gives
\begin{align*} \int \frac {1}{-a v +b}d v &= dt\\ -\frac {\ln \left (a v -b \right )}{a}&= t +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} -a v +b&= 0 \end{align*}
for \(v \left (t \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} v \left (t \right ) = \frac {b}{a} \end{align*}
Solving for \(v \left (t \right )\) gives
\begin{align*}
v \left (t \right ) &= \frac {{\mathrm e}^{-c_1 a -t a}+b}{a} \\
\end{align*}
The substitution \(v = p^{1-n}\) is now used to convert the above solution back to \(p\) which
results in
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that
satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
If the above condition is satisfied,
then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know
now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not
satisfied then this method will not work and we have to now look for an integrating
factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial t}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an
integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial t} \right ) \\ &=1\left ( \left ( 2 b p -a\right ) - \left (0 \right ) \right ) \\ &=2 b p -a \end{align*}
Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a
second method to find an integrating factor. Let
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial t} - \frac {\partial M}{\partial p} \right ) \\ &=-\frac {1}{p \left (-b p +a \right )}\left ( \left ( 0\right ) - \left (2 b p -a \right ) \right ) \\ &=\frac {2 b p -a}{p \left (-b p +a \right )} \end{align*}
Since \(B\) does not depend on \(t\), it can be used to obtain an integrating factor. Let the
integrating factor be \(\mu \). Then
\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {2 b p -a}{p \left (-b p +a \right )}\mathop {\mathrm {d}p} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{-\ln \left (p \left (b p -a \right )\right ) } \\ &= \frac {1}{p \left (b p -a \right )} \end{align*}
\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\)
and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{p \left (b p -a \right )}\left (b \,p^{2}-a p\right ) \\ &= 1 \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{p \left (b p -a \right )}\left (1\right ) \\ &= \frac {1}{p \left (b p -a \right )} \end{align*}
So now a modified ODE is obtained from the original ODE which will be exact and can be
solved using the standard method. The modified ODE is
Where \(f(p)\) is used for the constant of integration since \(\phi \) is a function
of both \(t\) and \(p\). Taking derivative of equation (3) w.r.t \(p\) gives
Where \(c_1\) is constant of integration. Substituting result found above for \(f(p)\) into
equation (3) gives \(\phi \)
\[
\phi = t +\frac {\ln \left (b p -a \right )}{a}-\frac {\ln \left (p \right )}{a}+ c_1
\]
But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new
constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as
\[
c_1 = t +\frac {\ln \left (b p -a \right )}{a}-\frac {\ln \left (p \right )}{a}
\]
Solving for the constant of integration from initial conditions, the solution becomes
\begin{align*} t +\frac {\ln \left (b p-a \right )}{a}-\frac {\ln \left (p\right )}{a} = \frac {\operatorname {t0} a +\ln \left (\operatorname {p0} b -a \right )-\ln \left (\operatorname {p0} \right )}{a} \end{align*}
Solving for \(p\) gives
\begin{align*}
p &= \frac {a \operatorname {p0}}{-{\mathrm e}^{-t a +\operatorname {t0} a} b \operatorname {p0} +{\mathrm e}^{-t a +\operatorname {t0} a} a +\operatorname {p0} b} \\
\end{align*}
Summary of solutions found
\begin{align*}
p &= \frac {a \operatorname {p0}}{-{\mathrm e}^{-t a +\operatorname {t0} a} b \operatorname {p0} +{\mathrm e}^{-t a +\operatorname {t0} a} a +\operatorname {p0} b} \\
\end{align*}
Substituting equations
(1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} b_{2}+\left (-b \,p^{2}+a p \right ) \left (b_{3}-a_{2}\right )-\left (-b \,p^{2}+a p \right )^{2} a_{3}-\left (-2 b p +a \right ) \left (p b_{3}+t b_{2}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-b^{2} p^{4} a_{3}+2 a b \,p^{3} a_{3}-a^{2} p^{2} a_{3}+b \,p^{2} a_{2}+b \,p^{2} b_{3}+2 b p t b_{2}-a p a_{2}-a t b_{2}+2 b p b_{1}-a b_{1}+b_{2} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -b^{2} p^{4} a_{3}+2 a b \,p^{3} a_{3}-a^{2} p^{2} a_{3}+b \,p^{2} a_{2}+b \,p^{2} b_{3}+2 b p t b_{2}-a p a_{2}-a t b_{2}+2 b p b_{1}-a b_{1}+b_{2} = 0
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{p, t\}\) in them.
\[
\{p, t\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{p, t\}\) in them
\[
\{p = v_{1}, t = v_{2}\}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} -b^{2} a_{3} v_{1}^{4}+2 a b a_{3} v_{1}^{3}-a^{2} a_{3} v_{1}^{2}+b a_{2} v_{1}^{2}+2 b b_{2} v_{1} v_{2}+b b_{3} v_{1}^{2}-a a_{2} v_{1}-a b_{2} v_{2}+2 b b_{1} v_{1}-a b_{1}+b_{2} = 0
\end{equation}
Collecting
the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}\}
\]
Equation (7E) now becomes
\begin{equation}
\tag{8E} -b^{2} a_{3} v_{1}^{4}+2 a b a_{3} v_{1}^{3}+\left (-a^{2} a_{3}+b a_{2}+b b_{3}\right ) v_{1}^{2}+2 b b_{2} v_{1} v_{2}+\left (-a a_{2}+2 b b_{1}\right ) v_{1}-a b_{2} v_{2}-a b_{1}+b_{2} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} -a b_{2}&=0\\ 2 b b_{2}&=0\\ -b^{2} a_{3}&=0\\ 2 a b a_{3}&=0\\ -a b_{1}+b_{2}&=0\\ -a a_{2}+2 b b_{1}&=0\\ -a^{2} a_{3}+b a_{2}+b b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( t,p\right ) \to \left ( R,S \right )\)
where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and
hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial t} + \eta \frac {\partial }{\partial p}\right ) S(t,p) = 1\). Starting with the first pair of ode’s in (1)
gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since
\(\xi =0\) then in this special case
\begin{align*} R = t \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{b \,p^{2}-a p}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (b p -a \right )}{a}-\frac {\ln \left (p \right )}{a} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
We now need to express the RHS as function of \(R\) only. This is done by solving for \(t,p\) in terms of
\(R,S\) from the result obtained earlier and simplifying. This gives
\begin{align*} \frac {dS}{dR} &= -1 \end{align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts
an ode, no matter how complicated it is, to one that can be solved by integration when the
ode is in the canonical coordiates \(R,S\).
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).